gcd of $x$ and $2$ in $\mathbb Z[x]$

In $\mathbb Z[x]$, $x$ and $2$ have gcd $1$. But they cannot be expressed as the linear combination of two polynomials.

Then assuming that $1=2f(x)+xg(x)$ we are supposed to arrive at a contradiction. If $g\equiv 0$ then $2f(x)=1$, but there is no such $f(x)$ that is multiplicative inverse of $2$ in $\mathbb Z[x]$. Is that correct? What is the contradiction when assumed $f$ is non-zero ?

Thanks for any help.

• What do you know about the constant term of $2 \cdot \sum_i a_i x^i + x \cdot \sum_i b_i x^i$? – Henry Swanson Aug 27 '15 at 5:46

$xg(x)$ has no constant term, so the constant term of $2f(x)+xg(x)$ is twice the constant term of $f$. But it must be equal to $1$, which is a contradiction because $1$ is not divisible by $2$.

Hint $\$ Evaluating $\,1 = 2 f(x) + x g(x)\,$ at $\,x=\color{#c00}0\,$ implies $1$ is even, contradiction.

Remark $\$ Responding to comments, I elaborate to explain the conceptual viewpoint behind this answer. The idea is similar to this answer, where we exploit the universal properties of the polynomial ring. But I'll avoid technical language in order to keep the exposition elementary.

The key idea in the hint is this. If the Bezout identity is true then $2$ and $x$ are coprime in $\,\Bbb Z[x]$ therefore $2$ is "universally" coprime to all $\,n\in\Bbb Z,\,$ contradiction (e.g. choose $\,n=\color{#c00}0)$.

Indeed, evaluating the Bezout identity at $\,x=n\,$ yields a Bezout identity in $\Bbb Z$ implying the coprimality of $2$ and $n$ (note the Bezout form is preserved since evaluation is a ring hom).

If we think of $\,\Bbb Z[x]$ as a ring obtained by adjoining an element $x$ that represents a universal or generic element of $\,\Bbb Z$ then we can think of the equations true in $\,\Bbb Z[x]$ as identities / laws of $\,\Bbb Z.\,$ For example the Bezout identity $1 = x^2 - (x\!-\!1)(x\!+\!1)$ can be viewed as a universal / generic proof that $\,n^2$ is coprime to $\,n\!\pm1\!\,$ for all $\,n.$

Thus, from this viewpoint, the hint amounts simply to disproving an identity by choosing an (obvious) counterexample. Moreover, the identity has (universal) number-theoretical content (about coprimality), which makes the search for a counterexample utterly trivial, i.e. searching for an integer that is not coprime to $2$.

The above remarks about universality are made precise when one studies universal algebra. In fact one can define / characterize polynomial rings (and free algebras) based on analogous ideas, e.g. see here..

See here for less trivial examples of proofs exploiting such universal ideas.